Algebra Examples

$(2, - 1)$ and $(4, 5)$

Step 1

Find the slope of the line between $(2, - 1)$ and $(4, 5)$ using $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ , which is the change of $y$ over the change of $x$ .

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Step 1.1

Slope is equal to the change in $y$ over the change in $x$ , or rise over run.

$m = \frac{change in y}{change in x}$

Step 1.2

The change in $x$ is equal to the difference in x-coordinates (also called run), and the change in $y$ is equal to the difference in y-coordinates (also called rise).

$m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$

Step 1.3

Substitute in the values of $x$ and $y$ into the equation to find the slope.

$m = \frac{5 - (- 1)}{4 - (2)}$

Step 1.4

Simplify.

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Step 1.4.1

Simplify the numerator.

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Step 1.4.1.1

Multiply $- 1$ by $- 1$ .

$m = \frac{5 + 1}{4 - (2)}$

Step 1.4.1.2

Add $5$ and $1$ .

$m = \frac{6}{4 - (2)}$

Step 1.4.2

Simplify the denominator.

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Step 1.4.2.1

Multiply $- 1$ by $2$ .

$m = \frac{6}{4 - 2}$

Step 1.4.2.2

Subtract $2$ from $4$ .

$m = \frac{6}{2}$

Step 1.4.3

Divide $6$ by $2$ .

$m = 3$

Step 2

Use the slope $3$ and a given point $(2, - 1)$ to substitute for $x_{1}$ and $y_{1}$ in the point-slope form $y - y_{1} = m (x - x_{1})$ , which is derived from the slope equation $m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}}$ .

$y - (- 1) = 3 \cdot (x - (2))$

Step 3

Simplify the equation and keep it in point-slope form.

$y + 1 = 3 \cdot (x - 2)$

Step 4

Solve for $y$ .

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Step 4.1

Simplify $3 \cdot (x - 2)$ .

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Step 4.1.1

Rewrite.

$y + 1 = 0 + 0 + 3 \cdot (x - 2)$

Step 4.1.2

Simplify by adding zeros.

$y + 1 = 3 \cdot (x - 2)$

Step 4.1.3

Apply the distributive property.

$y + 1 = 3 x + 3 \cdot - 2$

Step 4.1.4

Multiply $3$ by $- 2$ .

$y + 1 = 3 x - 6$

Step 4.2

Move all terms not containing $y$ to the right side of the equation.

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Step 4.2.1

Subtract $1$ from both sides of the equation.

$y = 3 x - 6 - 1$

Step 4.2.2

Subtract $1$ from $- 6$ .

$y = 3 x - 7$

Step 5